# Show that a given composite function is continuous

## Question

Let $$f : D \rightarrow \mathbb R$$ and $$g : E \rightarrow \mathbb R$$ be two uniformly continuous functions with $$f(D) \subseteq E$$. Show that the composite function $$g \circ f : D \rightarrow\mathbb R, x \mapsto g(f(x))$$ is uniformly continuous.

## Proof

We know that since $$f,g$$ are continuous

$$\Rightarrow\forall \space \epsilon_1\gt0 \space \space \exists \delta_1\gt0 \space$$ $$\forall x_1,y_1\in D:\space |x_1-y_1|\lt \delta_1\space\space \Rightarrow$$ $$|f(x_1)-f(y_1)|\lt \epsilon_1$$

and

$$\forall \space \epsilon_2\gt0 \space \space \exists \delta_2\gt0 \space$$ $$\forall x_2,y_2\in E:\space |x_2-y_2|\lt \delta_2\space\space\Rightarrow$$ $$|g(x_2)-g(y_2)|\lt \epsilon_2$$.

Since $$f(D)\subseteq E$$ $$\Rightarrow f(x_1),f(x_2)\in E$$.

So choosing $$x_2=f(x_1)$$, $$y_2=f(y_2)$$ and $$\epsilon_1=\delta_2$$

We get: $$\forall \space \epsilon_1\gt0 \space \space \exists \delta_1\gt0 \space$$ $$\forall x_1,y_1\in D:\space |x_1-y_1|\lt \delta_1\space\space \Rightarrow$$ $$|f(x_1)-f(y_1)|\lt \epsilon_1=\delta_2$$ $$\Rightarrow|g(f(x_1))-g(f(y_2))|\lt \epsilon_2$$.

Hence $$g \circ f : D \rightarrow\mathbb R, x \mapsto g(f(x))$$ is uniformly continuous.

It would be great if anyone could verify my solution. Any additional proofs or alternative angles would be greatly appreciated too.

What you had to show was for each $$\epsilon>0,$$ there exists $$\delta>0$$ such that $$|x_1-y_1|<\delta \implies |g(f(x_1)-g(f(y_1))|<\epsilon.$$

(Do you see how this is different from your last equation?)

Outline of proof: Let $$\epsilon>0,$$ since $$g$$ is uniformly continuous, there exists $$\delta'>0$$ such that $$|x_2-y_2|<\delta'\implies |g(x_2)-g(y_2)|<\epsilon.$$ Since $$f$$ is uniformly continuous, there exists $$\delta>0$$ such that $$|x_1-y_1|<\delta \implies|f(x_1)-f(x_2)|<\delta'\implies |g(f(x_1)-g(f(y_1))|<\epsilon.$$

• Okay, I think I understand what you're saying. Just updated the last few lines of the proof- is that any better? – George Cooper Apr 28 at 14:13
• @GeorgeCooper Yes, but you choose $\epsilon_1 = \delta_2$ not the other way round so be mindful while writing that. – Sahiba Arora Apr 28 at 14:20
• Okay, you very much for your help :) – George Cooper Apr 28 at 14:28

We know that since $$f,g$$ are continuous

$$\Rightarrow\forall \space \epsilon_1\gt0 \space \space \exists \delta_1\gt0 \space$$ $$\forall x_1,y_1\in D:\space |x_1-y_1|\lt \delta_1\space\space \Rightarrow$$ $$|f(x_1)-f(y_1)|\lt \epsilon_1$$

and

$$\forall \space \epsilon_2\gt0 \space \space \exists \delta_2\gt0 \space$$ $$\forall x_2,y_2\in E:\space |x_2-y_2|\lt \delta_2\space\space\Rightarrow$$ $$|g(x_2)-g(y_2)|\lt \epsilon_2$$.

Since $$f(D)\subseteq E$$ $$\Rightarrow f(x_1),f(x_2)\in E$$.

So choosing $$x_2=f(x_1)$$, $$y_2=f(y_2)$$ and $$\epsilon_1=\delta_2$$

We get: $$\forall \space \epsilon_1\gt0 \space \space \exists \delta_1\gt0 \space$$ $$\forall x_1,y_1\in D:\space |x_1-y_1|\lt \delta_1\space\space \Rightarrow$$ $$|f(x_1)-f(y_1)|\lt \epsilon_1=\delta_2$$ $$\Rightarrow|g(f(x_1))-g(f(y_2))|\lt \epsilon_2$$.

Hence $$g \circ f : D \rightarrow\mathbb R, x \mapsto g(f(x))$$ is uniformly continuous.